Solving $\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$

Where do I start to solve a equation for x like the one below?

$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$

After squaring it, it's too complicated; but there's nothing to factor or to expand?

Ideas?

Solution 1:

Divide $$(7x-4)-(7x-5)=(4x-1)-(4x-2)\ \ \ \ \ (1)$$ by the original equation, which is $$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}\ \ \ \ \ (2)$$ You should get:$$\sqrt{7x-4}+\sqrt{7x-5}=\sqrt{4x-1}+\sqrt{4x-2}\ \ \ \ \ (3)$$Now add (2) and (3), you get: \begin{align} 2\sqrt{7x-4}&=2\sqrt{4x-1} \\ \\ 7x-4&=4x-1 \\ \\ x&=1 \end{align}

Solution 2:

Note that

$$ \sqrt{7x-4} - \sqrt{7x-5} = \frac{(7x-4)-(7x-5)}{\sqrt{7x-4} + \sqrt{7x-5}} = \frac{1}{\sqrt{7x-4} + \sqrt{7x-5}} $$

and likewise

$$ \sqrt{4x-1} - \sqrt{4x-2} = \frac{1}{\sqrt{4x-1} + \sqrt{4x-2}}. $$

Plugging these to our equation, we obtain

$$ \sqrt{7x-4} + \sqrt{7x-5} = \sqrt{4x-1} + \sqrt{4x-2}. $$

Thus we have

$$ \sqrt{7x-4} = \sqrt{4x-1} \quad \text{and} \quad \sqrt{7x-5} = \sqrt{4x-2}, $$

yielding $x = 1$.

Solution 3:

From $$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$ multiply both sides by $$(\sqrt{7x-4}+\sqrt{7x-5})\times(\sqrt{4x-1}+\sqrt{4x-2})$$ obtaining $$\sqrt{4x-1}+\sqrt{4x-2} =\sqrt{7x-4}+\sqrt{7x-5}$$ so that $$2\sqrt{4x-1}=2\sqrt{7x-4}$$

And after dividing by $2$ and squaring, obtain $x=1$.

Note: this underlies he other answers so far given, but this approach is a form of "multiplying by the conjugate" which comes in handy in a variety of situations.

Solution 4:

$$\sqrt{7x-4}-\sqrt{7x-5}=\sqrt{4x-1}-\sqrt{4x-2}$$

$$\implies \sqrt{7x-4}+\sqrt{4x-2}=\sqrt{4x-1}+\sqrt{7x-5} $$

Squaring we get, $$7x-4+4x-2+2\sqrt{(7x-4)(4x-2)}=4x-1+7x-5+2\sqrt{(4x-1)(7x-5)}$$

$$\implies \sqrt{(7x-4)(4x-2)}= \sqrt{(4x-1)(7x-5)}$$

$$\text{Squaring we get, } (7x-4)(4x-2)=(4x-1)(7x-5)$$

$$\text{ On simplification we get, 8-(16+14)x=5-x(20+7)} $$

$$\implies x=1,\text{ which evidently satisfies the given equation. }$$

Generalization:

If $$\sqrt{ax+b}-\sqrt{ax+c}=\sqrt{dx+e}-\sqrt{dx+f} \text{ where }b-c=e-f$$

$$\implies \sqrt{ax+b}+\sqrt{dx+f}=\sqrt{dx+e}+\sqrt{ax+c} $$

Squaring we get, $$ax+b+dx+f+2\sqrt{(ax+b)(dx+f)}=dx+e+ax+c+2\sqrt{(dx+e)(ax+c)} $$

$$\implies \sqrt{(ax+b)(dx+f)}=\sqrt{(dx+e)(ax+c)}\text{ as } b-c=e-f\iff b+f=c+e $$

$$\text{Squaring we get, } (ax+b)(dx+f)=(dx+e)(ax+c)$$

$$\implies x(bd+af-cd-ae)=ec-bf $$

Solution 5:

Hint: below let $\, a,b,c,d\, =\, \sqrt{7x\!-\!4},\ \sqrt{7x\!-\!5},\ \sqrt{4x\!-\!1},\ \sqrt{4x\!-\!2}$

$$\begin{eqnarray} & a^2-b^2 &=\,& c^2-d^2 \ne 0 \\ & a-b &=\,& c-d \\ \Rightarrow & a+b &=\,& c+d\quad \text{via quotient of above two} \\ \Rightarrow & \qquad 2a &=\,& 2c\qquad\ \text{via sum of above two} \\ \Rightarrow &\qquad a &=\,& c\,\ \Rightarrow\ b=d \\ \Rightarrow &\qquad a^2 &=\,& c^2 \Rightarrow\ b^2\!=d^2 \end{eqnarray}$$